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If 0.1800 g of impure soda ash (Na2CO3) is titrated with 15.66 mL of 0.1082 M HCl,

If 0.1800 g of impure soda ash (Na2CO3) is titrated with 15.66 mL of 0.1082 M HCl, what is the percent purity of the soda ash?
Na2CO3(aq) + 2 HCl(aq) ? 2 NaCl(aq) + H2O(mc048-1.jpg) + CO2(g)
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Answer #2

moles HCl = 0.01566 L x 0.1082 M=0.001694
the balanced equation is

Na2CO3 + 2 HCl = H2O + CO2 + 2 NaCl

moles Na2CO3 = 0.001694/2=0.0008472
mass Na2CO3 = 0.0008472 mol x 105.9876 g/mol=0.08979 g
% = 0.08979 x 100/ 0.1800 =49.88

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